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Re: If x and y are integers, is x^7 < 6^y ? [#permalink]
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Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and independent equations ensures a solution.

If x and y are integers, is x^7 < 6^y ?

(1) x^3 = -125

(2) y^2 = 36

In the original condition, there are 2 variables (x,y), and 2 equations from the 2 conditions; this gives a high chance (C) will be our answer.
If we combine the 2 conditions,
x=-5, y=-6,6, which we can see that from x^7=(-3)^7 is already a negative number, making the conditions sufficient and the answer (C).
However this is a integer question, one of key question, so if we apply common mistakes type 4(A),
from condition 1, the answer becomes 'yes' when x=-5, as (-5)^7<0. This is sufficient.
from condition 2, the answer becomes 'no' when x=10 and y=6, but 'yes' when x=1, y=6. Therefore this is insufficient, and the answer is (A).
This is the type of question tested nowadays...

For cases where we need 2 more equation, such as original conditions with “2 variables”, or “3 variables and 1 equation”, or “4 variables and 2 equations”, we have 1 equation each in both 1) and 2). Therefore, there is 70% chance that C is the answer, while E has 25% chance. These two are the majority. In case of common mistake type 3,4, the answer may be from A, B or D but there is only 5% chance. Since C is most likely to be the answer using 1) and 2) separately according to DS definition (It saves us time). Obviously there may be cases where the answer is A, B, D or E.
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Re: If x and y are integers, is x^7 < 6^y ? [#permalink]
Bunuel wrote:
If x and y are integers, is x^7 < 6^y ?
(1) x^3 = -125
(2) y^2 = 36
Kudos for a correct solution.

Hi Bunuel,

OA is revealed. Also some good answers have been produced.

However no solution is rewarded :wink:

Just wanna know are we still looking for a correct solution?
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Re: If x and y are integers, is x^7 < 6^y ? [#permalink]
umm.. relishing first Kudos from the BOSS :-D

feeling legendary.
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Re: If x and y are integers, is x^7 < 6^y ? [#permalink]
Bunuel wrote:
If x and y are integers, is x^7 < 6^y ?

(1) x^3 = -125

(2) y^2 = 36


Kudos for a correct solution.


oh yeah! what a nice question!!!
1. if x^3 is negative, then x is negative, and x^7 is negative. regardless of the sign for y, x^7 will always be smaller than 6^y. sufficient.
2. y=6 or y=-6. but what about x?

A
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Re: If x and y are integers, is x^7 < 6^y ? [#permalink]
Prompt analysis
x and y are integers

Superset
The answer will be either yes or no

Translation
In order to find the answer we need
1# exact value of x and y
2# range of x and y

Statement analysis
St 1: x^3 =-125 => x = -5, no idea about the value of y. INSUFFICIENT
St 2: y^2 = 36 => y = 36 or -36. No idea about the value of x. INSUFFICIENT

St 1& St 2: x =-5 and y =6 or -6. -5^7 will be negative where as 6^6 or 6^-6 will be postive. The condition will hold true for both the cases. ANSWER

Option C
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Re: If x and y are integers, is x^7 < 6^y ? [#permalink]
Bunuel wrote:
If x and y are integers, is x^7 < 6^y ?

(1) x^3 = -125

(2) y^2 = 36


Kudos for a correct solution.


In order to solve this question we simply need to know whether x is negative or positive- because if x is negative then it doesn't matter whether y is negative or positive because both answers would be positive ( any negative exponent just results in a fraction)

Statement 1

x^3 = -125
x=-5

Sufficient

Statement 2

y^2 =36
y = -6 or 6

we don't know what the value of x is

insufficient
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Re: If x and y are integers, is x^7 < 6^y ? [#permalink]
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